Chapter 16
The Giancoli onLine Tutor

The Electric Charge and Electric Field

I. Coulomb's Law

Similarities and differences between the Electric Force and the Gravitational Force.


 Electric Force Gravitational Force Comment
1/r2 Yes Yes Both Forces are long-range forces
Strength strong weak
Atractive Yes Yes unlike charges attract
Repulsive Yes No like charges repel

To work a simple problem using Coulombs Law ignore the sign of the charges and just plug the charges and separation distance into the equation and evaluate it. This gives you the magnitude of the force. Determine the direction by remembering that 'like charges repel' and 'unlike charges attract' and that the force is along a line drawn between their centers.

In the following tutorials it is important that you understand the reason for the correct answers you discover. If you don't understand the reason ask your Professor.

Interactive Examples

In the figure:


1) If the blue and red charges are both positive, what is the direction of the force on the red charge? (The results will be in the window next to the selection list.)

2) If the blue and red charges are both positive, what is the direction of the force on the blue charge?


3) If the blue is positive and red charge is negative, what is the direction of the force on the blue charge?


4) If the blue is positive and red charge is negative, what is the direction of the force on the red charge?


5) If the blue charge is 1/ 3 mC and the red charge is 1/3 mC , the force will be exactly 1 N (using k=9x109) . If the distance is doubled the new force will be?


6) If the blue charge is 1/3 mC and the red charge is 1/3 mC , the force will be exactly 1 N . If one charge is doubled the new force will be?


7) If the blue charge is 1/3 mC and the red charge is 1/3 mC , the force will be exactly 1 N. If both charges are doubled the new force will be?


8) If the blue charge is 1/3 mC and the red charge is 1/3 mC , the force will be exactly 1 N. If both charges are doubled and the distance is, also, doubled the new force will be?


Using Vectors with Coulombs Law

Problems with more than two charges requires the use of vectors. If you are not comfortable with this please review the material in the chapter3 hints.

Interactive Activity

In this Figure the red, blue and green charges are each 1/3 mC while the yellow charge is 1/3 mC. Thus, as above, the magnitude of the force that the red and green charges exerts on the yellow charge has a magnitude of 1 N. The blue and green charges are negative, as shown, and the red and yellow charges are positive. Use this information to answer the folllowing questions.


1) What is the magnitude of the force the blue charge exerts on the yellow charge?
2) Which vector represents the direction of the force green charge exerts on the yellow charge?
3) Which vector represents the direction of the force blue charge exerts on the yellow charge?
4) Which vector represents the direction of the force red charge exerts on the yellow charge?
To answer the following questions use the following data, as appropriate:

All angles measured counter-clockwise from EAST.
sin(30o)=.50 cos(30o)=.87
sin(60o)=.87 cos(60o)=.50
sin(45o)=.71 cos(45o)=.71
sin(37o)=.60 cos(37o)=.80
sin(53o)=.80 cos(53o)=.60
sin(0o )=0.0 cos(0o )=1.0
sin(90o)=1.0 cos(90o)=0.0
sin(180o )=0.0 cos(180o )=-1.0
sin(225o)=-.71 cos(225o)=-.71
sin(270o)=-1.0 cos(270o)=0.0


Interactive Example

Enter the data below, after entering a number click outside of the box each time. If you made a mistake you will be coached. If nothing happens you got it right! Congratulations. Complete the following table (see Chapter 3 hints if you dont understand what to do).

Force X-Component Y-Component
Green (E)
Red (C)
Blue (F)
TOTALS
FEEDBACK:

Now that you have successfully completed the table; you have found the X and Y components of the total force on the yellow charge. Scroll down to see the results!







The Fx and Fy components are shown in the figure along with the resulting force vector F.

To find the magnitude of F use the Pythagorean Theorem.

|F|=SQRT(Fx2+Fy2)=SQRT(1.3552+.6452)= 1.5

Finding the angle is more problematic, using

q = tan-1(Fy/Fx)= tan-1(-1.355/.645)= tan-1(-2.1)

produces an answer of -64.5o (the angle defined by the blue line in the figure, clockwise angles are considered negative). The reason for this is that the inverse tangent function is double valued. It cannot distinguish between

tan-1(-1.355/.645) [second quadrant vector]

and

tan-1(.645/-1.355) [fourth quadrant vector]

As a result it chooses the smallest angle. All second quadrant vectors will produce results in the fourth quadrant. It is up to you to recognize this and add 180o to the result to obtain the correct answer. Similarly, all third quadrant results will be mirrored in the first quadrant. Again, you must add 180o. Finally, if you get a result that is actually in the fourth quadrant, convert it to the correct positive angle by adding 360o.

II. The Electric Field


The idea of the Electric Field is also introduced in this chapter. The Electric Field at a point in space is defined at a point in space as the force on a positive test charge, q, at that point divided by the charge,

E=F/q, as a result the unit of the Electric Field is N/C. The direction of the field will be the directio a positive charge would move if placed at that point. Thus it would be away from a simple positive charge and towards a negative charge. In terms of problem solving there are two important cases:


Point Charges


The Electric field is away from the positive charge and towards a negative charge. It is calculated just like the Electric Force with the second charge equal to 1 (without units).



Problems involving the Electric Field and point charges are worked very much like the ElectricForce problems. The adjoining figure shows an arangement of three charges and a point x where the field is to be calculated.

In the above Figure the red, blue and green charges are each 1/3 nC . Thus the magnitude of the Electric Field that the red and green charges produce at point x has a magnitude of 1 N/C. The blue and green charges are negative, as shown, and the red charge is positive. Use this information to answer the following questions!.

Interactive Examples

1) What is the magnitude of the Electric Field from the blue charge at the point x?
2) Which vector represents the direction of the Electric Field at x from the green charge ?
3) Which vector represents the direction of the Electric Field at x from the blue charge ?
4) Which vector represents the direction of the Electric Field at x from the red charge ?
To complete the problem you repeat what you did with the Electric Force, namely,
  1. Construct the component table
  2. Use the Pythagorean Theorem
  3. Use the inverse tangent


To find the Force on a charge, q, at this point use,

F=qE,

where E is your result from above. The direction of the force on a positive charge will be in the direction of the field and, conversely, for a negative charge, it will be opposite to the field.


The Capacitor (two oppositely, but equally, charged metal plates separated by a small distance)


In the next chapter we will learn how to calculate E inside a capacitor, but for the problems in this chapter E will normally be a given. You will probably be given a charge, q, also, from this you can calculate the force using. The Electric Field in a capacitor is constant in both magnitude and direction (except near the edges which we ignore). As a result most problems involving capacitors in this chapter reduce to constant force problems.

F=qE.

Once you have the force, you calculate the acceleration using Newton's Second Law. From there use the methods of Chapter 2.
© 1998 Carl Adler mailto:Carl@Image-ination.com